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4.4. Symmetric Novikov algebras
(2) If I is non-degenerate then so is I ⊥ and N = I ⊥
⊕ I ⊥ .
Proof.
(1) Since B(xN,I) = B(x,NI) = 0 and B(I,Nx) = B(IN,x) = 0, for all x ∈ I ⊥ then I ⊥ is
also an ideal of N. Let x ∈ II ⊥ , i.e. x = yz with y ∈ I,z ∈ I ⊥ then B(x,N) = B(yz,N) =
B(y,zN) = 0. Therefore x = 0. That means II ⊥ = {0}. Similarly, one gains I ⊥ I = {0}.
(2) Assume that I ⊥ is degenerate, that means there is a non-zero x ∈ I ⊥ such that B(x,I ⊥ ) =
0. Therefore, x ∈ I. However, B(x,I) = 0 since x ∈ I ⊥ so I is degenerate (that is a
contradiction). Hence, I ⊥ must be non-degenerate.
Let x ∈ I ∩ I⊥ then B(x,I) = 0. Since I is non-degenerate, one has x = 0. That means
I ∩ I⊥ = {0} and N = I ⊥
⊕ I⊥ .
Proposition 4.4.6. Let Z(N) = {x ∈ N | xy = yx, ∀ y ∈ N} the center of N and denote by
As(N) = {x ∈ N | (x,y,z) = 0, ∀ y,z ∈ N}. One has
(1) If N is a Novikov algebra then Z(N) ⊂ N(N) where N(N) = {x ∈ N | (x,y,z) = (y,x,z) =
(y,z,x) = 0, ∀ y,z ∈ N} is the nucleus of N (see Definition 4.1.11 (3)). Moreover, if N is
also commutative then N(N) = N = As(N) (that means N is an associative algebra).
(2) If (N,B) is a symmetric Novikov algebra then
Proof.
(i) Z(N) = [g(N),g(N)] ⊥
(ii) N(N) = As(N) = (N,N,N) ⊥ .
(iii) LAnn(N) = RAnn(N) = Ann(N) = (N 2 ) ⊥ .
(1) Assume that x is an element in Z(N). For all y,z ∈ N, one has:
(x,y,z) = (xy)z − x(yz) = (yx)z − x(yz) = (yz)x − x(yz) = 0.
By (III), we also have (y,x,z) = 0. Moreover, since
(y,z,x) = (yz)x − y(zx) = x(yz) − y(xz)
= (xy)z − (x,y,z) − (yx)z + (y,x,z) = (xy)z − (yx)z = 0
one obtains Z(N) ⊂ N(N). If N is commutative then Z(N) = N. Certainly, Z(N) =
N(N) = N.
(2) Let (N,B) be a symmetric Novikov algebra.
(i) It is obvious since g(N) is a quadratic Lie algebra and Z(N) is its center.
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